Question

An open tubular column with a diameter of 203 μm and a stationary phase thickness on the inner wall of 0.75 μm passes unretained solute through in 53 s. A particular solute has a retention time of 435 s.

a.) What is the partition coefficient for this solute?

K=_____

b.) What fraction of time does this solute spend in the stationary phase?

Ts/Ttotal=_____

Answer 1

solution:

b)fraction of time spent in stationary phase = time spent in stationary phase/total time spent

fraction of time spent in stationary phase = (435 – 53 s)/435 s = 0.878 = 87.8%

a)k = (435 – 53 s)/53 s = 7.2

K = C_s/C_m = k(V_m/V_s)

V_m = V_cylinder = π(d²/4)L = π[(0.203 mm)²/4)L = 0.0324 mm²L

V_s = V_tube = V_{outer cylinder} – V_{inner cylinder} = π[(0.203 mm)²/4]L – π{[(0.207 mm – 2(0.00075 mm)]²/4}L

Note: for a thin tube, V_tube = πtdL, where t = film thickness. This same equation also can be calculated from the first equation (V_s = V_tube = V_{outer cylinder} – V_{inner cylinder} ) since

V_tube = π[(0.203 mm)²/4]L – π[(0.203 mm)²/4]L + π(0.203 mm)(0.00075 mm)L –

π(0.0075 mm)²L. The first two terms cancel out and since 0.203 mm >> 0.00075 mm, the last term is inconsequential.

V_s = V_tube = π(0.203 mm)(0.00075 mm)L = 0.0005 mm²L

K = k(V_m/V_s) = 7.2(0.0324 mm²L/0.0005 mm²L) = 467 (note how L cancels out so the column length is not needed).